1. Pat Arnott, ATMS 749 Atmospheric Radiation Transfer Air: n r =1.00006, n i =1e -10 Substance: n r =1.67, n i =0.2 I0I0 0x Given flux I 0 incident on the air:substance boundary. Calculate the Flux Transmitted to Point X:

2. Pat Arnott, ATMS 749 Atmospheric Radiation Transfer Given flux I 0 incident on the air:substance boundary. Calculate the Flux Transmitted to Point X: Air: n r =1.00006, n i =1e -10 Substance: n r =1.67, n i =0.2 I0I0 0x Answer: 1 R 1-R=T Propagator from 0 to x.

3. Pat Arnott, ATMS 749 Atmospheric Radiation Transfer 0x What if we divide the substance into particles? Calculate the Flux Transmitted to Point X: Air: n r =1.00006, n i =1e -10 Substance: n r =1.67, n i =0.2 I0I0 N identical particles / volume v = particle volume a = average particle projected area.  ext = a Q ext = Single Particle Extinction Cross Section. Q ext =Extinction Efficiency.  ext =  abs +  sca,  ext =  ext (n r,n i, ) P ext = I 0  ext = Power (watts) removed by a single particle from I 0 by extinction.

4. Pat Arnott, ATMS 749 Atmospheric Radiation Transfer 0x What if we divide the substance into particles? Calculate the Flux Transmitted to Point X: Air: n r =1.00006, n i =1e -10 Substance: n r =1.67, n i =0.2 I0I0 N particles / volume v = particle volume a = average projected area for each particle.  ext = a Q ext = Particle Extinction Cross Section. Q ext =Extinction Efficiency.  ext =N  ext =Extinction Coefficient. Assume  sca = N  sca x <<1. (single scattering assumption). I(x)=I 0 exp(-  ext x) Otherwise, use multiple scattering theory (to be developed soon)

5. Pat Arnott, ATMS 749 Atmospheric Radiation Transfer Single Particle Perspective: Assume  ext ≈  abs,  sca ≈ 0 (particle size much less than the wavelength, deep in the Rayleigh range. Size parameter << 1.) absorption Gross, Special Purpose, ad-hoc Approximation:  abs = a[1-exp(-D eq /  )]. Let D eq =v/a.  = /(4  n i )=skin depth. Limits: D eq << , (1-e -small )≈small,  abs = 4  n i v/  D eq >> , (1-e -large )≈1,  abs = a. v = particle volume a = particle projected area D eq

6. Pat Arnott, ATMS 749 Atmospheric Radiation Transfer 0x What if we divide the substance into particles? Calculate the Flux Transmitted to Point X: Air: n r =1.00006, n i =1e -10 Substance: n r =1.67, n i =0.2 I0I0 N particles / volume v = particle volume a = average projected area for each particle.  abs = N  abs x. D eq << ,  abs = 4  n i v/  I(x)=I 0 exp(- 4  n i vNx/ ) vN=C=(Particle Volume)/Volume C=Concentration (e.g. ppm v ) I(x)=I 0 exp(- 4  n i xC/ )

7. Pat Arnott, ATMS 749 Atmospheric Radiation Transfer Air: n r =1.00006, n i =1e -10 Substance: n r =1.67, n i =0.2 I0I0 x 0 Compare (Assumes no particle scattering, dilute (C<<1), weak absorption). C=volumetric concentration.

8. Pat Arnott, ATMS 749 Atmospheric Radiation Transfer Homework: Compare Mie Theory for Spheres with the simple model for absorption below. Gross Special Purpose Approximation:  abs = a[1-exp(-D eq /  )]. Let D eq =v/a.  = /(4  n i )=skin depth. v=4  r 3 /3. a=average projected area=  r 2 for a sphere. D=2r. D eq =2D/3. Cases in a 3 matrices for fixed n r and variable D and n i : (calculate the percentage error of the model and Mie theory.) = 0.5 um. n r =1, n r =1.33, n r =1.5 D=0.01 um, 0.1 um, 1 um, 10 um. n i =0.001, n i =0.01, n i =0.1, n i =1.

9. Pat Arnott, ATMS 749 Atmospheric Radiation Transfer Table for Homework (one for each real refractive index, 1.0, 1.333, and 1.5). Fill each empty table with a percentage error as defined below. D (microns) n i 0.010.1110 0.001 0.01 0.1 1

10. Pat Arnott, ATMS 749 Atmospheric Radiation Transfer Air: n r =1.00006, n i =1e -10 Bulk Substance: n r =1.67, n i =0.2 I0I0 x 0 Compare: Bulk Substance, Gas, and Particles (Assumes no particle scattering, dilute (C<<1), weak absorption). C=volumetric concentration. Gas Particles